A series to prove $\frac{22}{7}-\pi>0$

Question

After T. Piezas answered Is there a series to show $22\pi^4>2143\,$? a natural question is

Is there a series that proves $\frac{22}{7}-\pi>0$?

One such series may be found combining linearly the series that arise from truncating $$\sum_{k=0}^\infty \frac{48}{(4k+3)(4k+5)(4k+7)(4k+9)} = \frac{16}{5}-\pi$$ to two and three terms, namely

$$\sum_{k=2}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{141616}{45045}-\pi$$ and $$\sum_{k=3}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{2406464}{765765}-\pi$$ Solving $$a\left(\frac{141616}{45045}-\pi\right)+b\left(\frac{2406464}{765765}-\pi\right)=\frac{22}{7}-\pi$$ for rational $a,b$ and some algebra manipulation yields the result

$$\frac{16}{21} \sum_{k=0}^\infty \frac{1008 k^2+6952 k+12625}{(4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21)}=\frac{22}{7}-\pi$$

It is interesting to note that the coefficients needed to multiply the two component series are both positive $$a=\frac{113}{7·8·9}$$ $$b=\frac{391}{7·8·9}$$

because the truncation points have been chosen so that

$$\frac{2406464}{765765}<\frac{22}{7}<\frac{141616}{45045}$$

This procedure yields a result that proves the claim with no need for further processing, and it is readily seen to prove $\frac{p}{q}-\pi>0$ for all fractions between $\pi$ and $\frac{16}{5}$.

Now, in the light of this equivalent form of Lehmer's formula $$\pi-3=\sum_{k=1}^\infty \frac{4!}{(4k+1)(4k+2)(4k+4)}$$

one may still ask

Q1 Is there a series that proves $\frac{22}{7}-\pi>0$ with constant numerator?

Q2 Is there a reason why $113$ is both the numerator of the $a$ coefficient and the denominator of the next convergent from above $\frac{355}{113}$?

Edit: A similar series with smaller coefficients may be obtained by applying the method above to $$\begin{align} \sum_{k=0}^\infty \frac{960}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13)} &= \frac{992}{315}-\pi \\ &= \frac{3·333-7}{3·106-3}-\pi \\ \end{align} $$ in order to obtain $$\sum_{k=0}^\infty \frac{96 (160 k^2+422 k+405)}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17)} = \frac{22}{7}-\pi$$

Q3 What is the relationship between $\frac{992}{315}$ and the third convergent to $\pi$ $\frac{333}{106}$?

Answer 1

Q1

Evaluating the following series $$\begin{align} &\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)} \\ &= \sum_{k=0}^\infty \left(\frac{1}{4k+5}-\frac{4}{4k+6}+\frac{5}{4k+7}-\frac{5}{4k+9}+\frac{4}{4k+10}-\frac{1}{4k+11}\right) \\ &= \sum_{k=0}^\infty \int_{0}^1\left(x^{4k+4}-4x^{4k+5}+5x^{4k+6}-5x^{4k+8}+4x^{4k+9}-x^{4k+10}\right)dx \\ &= \int_{0}^1 x^4\sum_{k=0}^\infty \left(x^{4k}-4x^{4k+1}+5x^{4k+2}-5x^{4k+4}+4x^{4k+5}-x^{4k+6}\right)dx \\ &= \int_{0}^1 x^4\frac{1-4x+5x^2-5x^4+4x^5-x^6}{1-x^4}dx \\ &= \int_{0}^1 x^4\frac{(1-x^2)(1-x)^4}{(1-x^2)(1+x^2)}dx=\int_{0}^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi \\ \end{align}$$ shows its connection with Dalzell's integral.

This may be rewritten as $$\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$

which appears in the 2009 document by Peter Bala New series for old functions http://oeis.org/A002117/a002117.pdf (formula 5.1) and shows that $\frac{22}{7}-\pi$ can be obtained by taking one term out of the summation in the series $$\sum_{k=0}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{10}{3}-\pi$$

Consecutive truncations yield the inequality

$$\pi...<\frac{141514}{45045}<\frac{10886}{3465}<\frac{22}{7}<\frac{10}{3}$$

Similar fractions, but now converging to $\pi$ from below, may be obtained from the series

$$\sum_{k=0}^\infty \frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)} = \pi-\frac{47}{15}$$

This yields

$$\frac{47}{15}<\frac{1979}{630}<\frac{141511}{45045}<\frac{9622853}{3063060}<...\pi$$

(See a similar inequality for $\log(2)$)

Correspondence between series and integrals

$$\sum_{k=n}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\int_0^1 \frac{x^{4n}(1-x)^4}{1+x^2}dx$$

$$\sum_{k=n}^\infty \frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)}=\int_0^1 \frac{x^{4n+2}(1-x)^4}{1+x^2}dx$$

Equivalent expressions

The general term for these series may be written in compact form using factorials, binomial coefficients or the Beta integral $B$ (see this comment by N. Elkies).

$$\begin{align} \frac{22}{7}-\pi &= 3840\sum_{k=1}^\infty \frac{(k+2)!(4k)!}{(4k+8)!k!} \\ \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{\displaystyle{k+2 \choose 2}}{\displaystyle{4k+8\choose 8}} \\ \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{k+1}{\displaystyle{4k+7\choose 7}} \\ \\ &= \frac{16}{21} \sum_{k=1}^\infty \frac{B(4k+1,8)}{B(k+1,2)} \end{align} $$

Interpretation of $\frac{22}{7}-\pi$

Similar series and approximations

If we use the Pochhammer symbol to express this series: $$\sum_{k=0}^\infty \frac{7!(k+1)}{(4k+1)_7}=\frac{7}{4}(10-3\pi)\approx 1$$

we can change the numbers to obtain variants such as

$$ \sum_{k=0}^\infty \frac{5!(k+1)}{(3k+1)_{5}} = \frac{5}{9}\left(2\sqrt{3}\pi-9\right)\approx 1, $$

$$\sum_{k=0}^\infty \frac{11! (k+1)}{(6 k+1)_{11}} = 231-\frac{4565 \pi}{36 \sqrt{3}}\approx 1$$

and $$\sum_{k=0}^\infty \frac{15!(k+1)}{(8k+1)_{15}}=\frac{15}{8}(1716-7(99\sqrt{2}-62)\pi)\approx 1$$

Given that all three series evaluate to almost 1, the following corresponding approximations are derived

$$\begin{align} \pi &=\frac{9\sqrt{3}}{5}+\sqrt{3}\int_0^1\frac{x^3(1-x)^2(1+x)}{1+x+x^2}dx\\ &\approx\frac{9\sqrt{3}}{5} \\ \pi &=\frac{1656\sqrt{3}}{913}- \frac{6\sqrt{3}}{83}\int_0^1 \frac{x^6(1-x)^8}{1+x^2+x^4} dx\\ &\approx\frac{1656\sqrt{3}}{913} \\ \pi &=\frac{1838 \left(62 + 99 \sqrt{2}\right)}{118185}-\frac{62+99\sqrt{2}}{15758}\int_0^1 \frac{x^8(1-x)^{12}}{1+x^2+x^4+x^6}dx\\ &\approx \frac{1838 \left(62 + 99 \sqrt{2}\right)}{118185} \end{align}$$

which give 1, 5 and 8 correct decimals respectively.

The fraction $\frac{1838}{118185}$ is the eighth convergent of $\frac{\pi}{62+99\sqrt{2}}$

Another series and integral for $\frac{22}{7}-\pi$

$$\begin{align} &\sum_{k=0}^\infty \frac{285120}{(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)(4k+13)(4k+14)} \\ &= \frac{1}{28}\int_{0}^1 \frac{x(1-x)^8(2+7x+2x^2)}{1+x^2}dx=\frac{22}{7}-\pi \\ \end{align}$$

Answer 2

Let $\sum_{k=0}^\infty a_n$ any series converging to $\pi$ and choose any series converging to $\frac{22}{7}$, for instance $\sum_{k=0}^\infty \left(\frac{15}{22}\right)^n$

No problem to show that $$\sum_{k=0}^\infty \left(\left(\frac{15}{22}\right)^n -a_n\right)=\frac{22}{7}-\pi\gt 0 $$

Answer 3

Proof that $\frac{22}{7}$ exceeds $\pi$.

$$0<\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi$$

Proof-

$$\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$$

$$=\int_0^1x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}dx$$

$$=\frac {x^7}{7}+\frac{2x^6}{3}+x^5-\frac{4x^3}{3}+4x-4\tan^{-1}(x)\vert_0^1$$

Now,by applying $\tan^{-1}1=45^\circ=\frac\pi4$ and substituting it in the integral and solving the integral yields $\frac{22}{7}-\pi$